100 people are waiting to board a plane. The first person’s ticket says Seat 1; the second person in line has a ticket that says Seat 2, and so on until the 100th person, whose ticket says Seat 100. The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible; if their seat is taken, they will randomly choose one of the remaining seats (after the first person, the second person takes a seat; after the second person, the third person takes a seat, and so on). What is the probability the 100th person sits in Seat 100?

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answer 1%
If the first person to sit down does not sit in seat 1, then in the end seat 100 will be taken before the 100th person boards.
However, if the first person to board sits in seat 1, then all the following passengers will go to their assigned seats.

exception: if there are more than 100 seats on the plane then the solution cannot be found

There is an elegant answer that doesn’t take too long to state. Try doing small cases first to get a feel for the problem. Imagine there are only 2 seats and apply your logic. Imagine only 3 seats. Only 4 seats. Do you see a pattern?

Actually, it’s much higher than 1 out of 98. HINT: Try to do some special cases. See what happens if there are 2 seats, or 3 seats, or 4 seats. These cases are small enough to analyze by brute force, and you might see a pattern.

For those stating 1/100 it is very easy to see that the probability is much higher. Consider that if the 1st man chooses his own seat (a 1/100 occurrence), you already have reached 1% chance of the 100th man sitting in his own. However consider the possibility that the 1st man chooses to sit in seat m, and the mth man chooses to sit in seat 1. In this case the 100th man still gets his seat which establishes the lower bound of this problem well above 1%. I’m not strong in probability so I’ll have to play around longer to find an exact solution myself.

I used ur way to count that. I stared from 2 seat, then goed 3 seat an so on.

I noticed one thing.
With one seat it is 1 out of 1 that 1st man gets seat nro. 1
With two seats probality is 1 out of 2 that second man gets seat nro. 2
With three seats it is 2 out of 6 that 3th man gets seat nro. 3
With four seats it is 6 out of 24 to 4th man gets seat nro. 4

so here is little more easily version.
with one seat it is 1/1
1 out of 2 is same as 1/2
2 out of 6 is same as 1/3
6 out of 24 is same as 1/4
so…
in hundred it will be 1/100
so it is 1% change with this caculation

if that is not correct please tell me why this caculation does not work. (cause my caculator says that this is correct)

You’ve read the problem incorrectly. All 3! = 6 permutations are NOT possible. Remember every person after person 1 MUST sit in their seat if available. Thus if person 1 sits in seat 1, then person 2 MUST sit in seat 2. Your case of 1 3 2 is thus not possible; other cases may be impossible as well.

100 seats that can have a 100! different arrangments. 100 factorial = 9.33262154 × 101^57 that includes the 100th being in his 100th seat.
Since each is an independent event, then they are all equally likely to happen.
P( of the 100th seat to being empty for the 100th passenger ) = 99*99/ 9.33262154 × 101^57

You have to calculate the odds of each person having the possibility of choice and then multiply this probability by the odds that that person randomly selects seat 100. Once you have calculated these sums for all 99 people, then you can sum them, and this is the probability the final seat is taken.

For the first person the odds of having a choice are 100% and the odds of selecting seat 100 is 1/100. This gives person 1 a 1% chance of filling seat 100.

Person 2 will only have the option of choice if person 1 selects seat 2. This will happen 1 in 100 times and person 2 will select seat seat 100 1 in 99 times. The odds of person 2 selecting seat 100 in 1 in 99 or 1/9900

Person 3 will have the option of choice if person 1 or person 2 take seat 3. The odds of person 1 doing this is 1 in 100. The odds of person 2 doing this is 1 in 100 times 1 in 99. The justification behind this is that person 2 can only take seat 3 if person 1 takes seat 2 and then person 2 randomly selects seat 3. We can then sum the different probabilities of person 3 having a choice and multiply this sum by the odds person 3 select seat 100, or 1 in 98.

This process becomes very long and tedious. I used excel and mapped the whole thing out in a 100×100 grid and found the answer to be approximately

“The first person ignores the fact that his ticket says Seat 1, and randomly chooses one of the hundred seats (note: he might randomly choose to sit in Seat 1). From this point on, the next 98 people will always sit in their assigned seats if possible”

Well, I think it’s fine as is. Remember there are 100 people and 100 seats. After the first person, after the next 98 then only one person remains, and they MUST sit in whatever seat is free. If, however, you want to view it as the next 99 that’s fine.

It’s not 1 — that would mean the person ALWAYS gets their seat. Try special cases. Imagine that there are just 2 people and find the answer. Then do just 3 people. Then 4. You should hopefully see an interesting pattern.

That’s not the answer — also, the sum of the reciprocals is not the reciprocal of the sum. Try the following hint first: try to solve the problem when there are just 2 people, then just 3, then just 4. You should see something remarkable about these three answers. Do these cases out by brute force, seeing where people sit….

The first guy has 99 options to choose from, and the rest of the people have 99! Ways to arrange themselves. The last guy chooses his seat 98 of the times (When the first guy doesn’t choose the 100th seat). There are then 98 times 98! Factorial ways the others can arrange themselves. 98×98!/99×99! Gives us the answer

first person takes seat 1 33% of the time, seat 2 33% of the time, seat 3 33% of the time

if took seat 1, then person 2 takes seat 2 and then 3 takes 3
if took seat 2 then 50% of the time person 2 takes seat 1 and
50% of the time person 2 takes seat 3
if took seat 3 then doesn’t matter as can’t win

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